I have imputed missing values using Amelia thereby creating 5 multiply imputed datasets. Now, I would like to split this multi-dataset, e.g. one set for year => 1990 and one set for year =<1990. Any ideas how I can do so? Many thanks!
data(freetrade)
freetrade$year #splitting variable
#Imputation of missing data
a.out <- amelia(freetrade, m=5, ts="year", cs="country")
#split of created dataset?
Amelia 返回一个包含数据框列表的对象（对于每个插补）。您可以使用 来查看此对象的结构str()。
> library(Amelia)
> data(freetrade)
>
> a.out <- amelia(freetrade, m=5, ts="year", cs="country")
-- Imputation 1 --
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
-- Imputation 2 --
1 2 3 4 5 6 7 8 9 10 11 12 13
-- Imputation 3 --
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
-- Imputation 4 --
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
-- Imputation 5 --
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
> str(a.out)
List of 12
$ imputations:List of 5
..$ imp1:'data.frame': 171 obs. of 10 variables:
.. ..$ year : int [1:171] 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 ...
.. ..$ country : chr [1:171] "SriLanka" "SriLanka" "SriLanka" "SriLanka" ...
.. ..$ tariff : num [1:171] 30.6 22.4 41.3 26.8 31 ...
.. ..$ polity : num [1:171] 6 5 5 5 5 5 5 5 5 5 ...
.. ..$ pop : num [1:171] 14988000 15189000 15417000 15599000 15837000 ...
.. ..$ gdp.pc : num [1:171] 461 474 489 508 526 ...
.. ..$ intresmi: num [1:171] 1.94 1.96 1.66 2.8 2.26 ...
.. ..$ signed : num [1:171] 0 0 1 0 0 0 0 1 0 0 ...
.. ..$ fiveop : num [1:171] 12.4 12.5 12.3 12.3 12.3 ...
.. ..$ usheg : num [1:171] 0.259 0.256 0.266 0.299 0.295 ...
..$ imp2:'data.frame': 171 obs. of 10 variables:
.. ..$ year : int [1:171] 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 ...
.. ..$ country : chr [1:171] "SriLanka" "SriLanka" "SriLanka" "SriLanka" ...
.. ..$ tariff : num [1:171] 33.6 59.7 41.3 18.2 31 ...
.. ..$ polity : num [1:171] 6 5 5 5 5 5 5 5 5 5 ...
.. ..$ pop : num [1:171] 14988000 15189000 15417000 15599000 15837000 ...
.. ..$ gdp.pc : num [1:171] 461 474 489 508 526 ...
.. ..$ intresmi: num [1:171] 1.94 1.96 1.66 2.8 2.26 ...
.. ..$ signed : num [1:171] 0 0 1 0 0 0 0 1 0 0 ...
.. ..$ fiveop : num [1:171] 12.4 12.5 12.3 12.3 12.3 ...
.. ..$ usheg : num [1:171] 0.259 0.256 0.266 0.299 0.295 ...
..$ imp3:'data.frame': 171 obs. of 10 variables:
.. ..$ year : int [1:171] 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 ...
.. ..$ country : chr [1:171] "SriLanka" "SriLanka" "SriLanka" "SriLanka" ...
.. ..$ tariff : num [1:171] 48.5 32.9 41.3 47.2 31 ...
.. ..$ polity : num [1:171] 6 5 5 5 5 5 5 5 5 5 ...
.. ..$ pop : num [1:171] 14988000 15189000 15417000 15599000 15837000 ...
.. ..$ gdp.pc : num [1:171] 461 474 489 508 526 ...
.. ..$ intresmi: num [1:171] 1.94 1.96 1.66 2.8 2.26 ...
.. ..$ signed : num [1:171] 0 0 1 0 0 0 0 1 0 0 ...
.. ..$ fiveop : num [1:171] 12.4 12.5 12.3 12.3 12.3 ...
.. ..$ usheg : num [1:171] 0.259 0.256 0.266 0.299 0.295 ...
..$ imp4:'data.frame': 171 obs. of 10 variables:
.. ..$ year : int [1:171] 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 ...
.. ..$ country : chr [1:171] "SriLanka" "SriLanka" "SriLanka" "SriLanka" ...
.. ..$ tariff : num [1:171] 18.4 45.5 41.3 16.9 31 ...
.. ..$ polity : num [1:171] 6 5 5 5 5 5 5 5 5 5 ...
.. ..$ pop : num [1:171] 14988000 15189000 15417000 15599000 15837000 ...
.. ..$ gdp.pc : num [1:171] 461 474 489 508 526 ...
.. ..$ intresmi: num [1:171] 1.94 1.96 1.66 2.8 2.26 ...
.. ..$ signed : num [1:171] 0 0 1 0 0 0 0 1 0 0 ...
.. ..$ fiveop : num [1:171] 12.4 12.5 12.3 12.3 12.3 ...
.. ..$ usheg : num [1:171] 0.259 0.256 0.266 0.299 0.295 ...
..$ imp5:'data.frame': 171 obs. of 10 variables:
.. ..$ year : int [1:171] 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 ...
.. ..$ country : chr [1:171] "SriLanka" "SriLanka" "SriLanka" "SriLanka" ...
.. ..$ tariff : num [1:171] 15.3 44.4 41.3 40.1 31 ...
.. ..$ polity : num [1:171] 6 5 5 5 5 5 5 5 5 5 ...
.. ..$ pop : num [1:171] 14988000 15189000 15417000 15599000 15837000 ...
.. ..$ gdp.pc : num [1:171] 461 474 489 508 526 ...
.. ..$ intresmi: num [1:171] 1.94 1.96 1.66 2.8 2.26 ...
.. ..$ signed : num [1:171] 0 0 1 0 0 0 0 1 0 0 ...
.. ..$ fiveop : num [1:171] 12.4 12.5 12.3 12.3 12.3 ...
.. ..$ usheg : num [1:171] 0.259 0.256 0.266 0.299 0.295 ...
..- attr(*, "class")= chr [1:2] "mi" "list"
$ m : num 5
$ missMatrix : logi [1:171, 1:10] FALSE FALSE FALSE FALSE FALSE FALSE ...
..- attr(*, "dimnames")=List of 2
.. ..$ : NULL
.. ..$ : chr [1:10] "year" "country" "tariff" "polity" ...
$ overvalues : NULL
$ theta : num [1:9, 1:9, 1:5] -1 -0.08456 -0.03404 -0.00193 0.06483 ...
$ mu : num [1:8, 1:5] -0.08456 -0.03404 -0.00193 0.06483 -0.11178 ...
$ covMatrices: num [1:8, 1:8, 1:5] 0.7881 -0.1869 -0.0531 0.2121 -0.0819 ...
$ code : num 1
$ message : chr "Normal EM convergence."
$ iterHist :List of 5
..$ : num [1:15, 1:3] 44 34 25 28 26 25 24 22 20 14 ...
..$ : num [1:13, 1:3] 44 27 24 22 22 21 18 17 14 11 ...
..$ : num [1:19, 1:3] 44 34 29 27 26 26 25 24 23 21 ...
..$ : num [1:15, 1:3] 44 34 27 28 23 24 23 23 19 19 ...
..$ : num [1:20, 1:3] 44 32 30 27 24 23 23 23 23 21 ...
$ arguments :List of 22
..$ idvars : NULL
..$ logs : NULL
..$ ts : num 1
..$ cs : num 2
..$ empri : NULL
..$ tolerance : num 1e-04
..$ polytime : NULL
..$ splinetime : NULL
..$ lags : NULL
..$ leads : NULL
..$ intercs : logi FALSE
..$ sqrts : NULL
..$ lgstc : NULL
..$ noms : NULL
..$ ords : NULL
..$ priors : NULL
..$ autopri : num 0.05
..$ bounds : NULL
..$ max.resample: num 100
..$ startvals : num 0
..$ overimp : NULL
..$ emburn : num [1:2] 0 0
..- attr(*, "class")= chr [1:2] "ameliaArgs" "list"
$ orig.vars : chr [1:10] "year" "country" "tariff" "polity" ...
- attr(*, "class")= chr "amelia"
从这里您可以看到 a.out 对象的“估算”元素包含您的数据框，因此您可以从那里引用您的每个估算。例如a.out$imputations[[1]]$year，将为您提供第一次估算的年份。如果您喜欢在每个插补中执行此操作，则可以使用应用函数或循环来执行此操作。为了说明这一点，请考虑：
> sapply(a.out$imputations,function(x) head(x$year))
imp1 imp2 imp3 imp4 imp5
[1,] 1981 1981 1981 1981 1981
[2,] 1982 1982 1982 1982 1982
[3,] 1983 1983 1983 1983 1983
[4,] 1984 1984 1984 1984 1984
[5,] 1985 1985 1985 1985 1985
[6,] 1986 1986 1986 1986 1986
编辑：我刚刚重新阅读了您的问题，我发现您实际上正在寻找更具体的东西。您可以将上面的内容应用到使每个数据帧的子集执行类似lapply(a.out$imputations,function(x) x[x$year > 1990,]). 我不确定你想如何组合这些估算的数据集（按年分割大于/小于 1990 年），但如果你只想将所有行附加在一起rbind()就可以了（如果不让我知道你会如何喜欢，我可能会推荐一个解决方案）：
> df1 <- do.call(rbind,lapply(a.out$imputations,function(x) x[x$year > 1990,]))
> df2 <- do.call(rbind,lapply(a.out$imputations,function(x) x[x$year < 1990,]))
> head(df1)
year country tariff polity pop gdp.pc intresmi signed fiveop usheg
imp1.11 1991 SriLanka 26.9000 5 17247000 597.6987 2.285213 1.000000 12.8 0.2589872
imp1.12 1992 SriLanka 25.0000 5 17405000 618.3329 2.877877 0.515665 13.1 0.2623017
imp1.13 1993 SriLanka 24.2000 5 17628420 652.6205 4.280361 0.000000 13.2 0.2812928
imp1.14 1994 SriLanka 26.0000 5 17865000 680.0408 4.389912 0.000000 13.2 0.2783585
imp1.15 1995 SriLanka 20.0000 5 18112000 707.6591 3.995919 0.000000 13.2 0.2627195
imp1.16 1996 SriLanka 20.5646 5 18300000 727.0039 3.676763 0.000000 13.2 0.2681700
> head(df2)
year country tariff polity pop gdp.pc intresmi signed fiveop usheg
imp1.1 1981 SriLanka 30.56693 6 14988000 461.0236 1.937347 0 12.4 0.2593112
imp1.2 1982 SriLanka 22.39382 5 15189000 473.7634 1.964430 0 12.5 0.2558008
imp1.3 1983 SriLanka 41.30000 5 15417000 489.2266 1.663936 1 12.3 0.2655022
imp1.4 1984 SriLanka 26.81580 5 15599000 508.1739 2.797462 0 12.3 0.2988009
imp1.5 1985 SriLanka 31.00000 5 15837000 525.5609 2.259116 0 12.3 0.2952431
imp1.6 1986 SriLanka 17.76314 5 16117000 538.9237 1.832549 0 12.5 0.2886563